Thread: Hyperbolic Integral Problem

How do I integrate this:

(x^2)sinh[mx] dx

My Calc I professor skipped over this, saying we wouldn't need it. Turns out, we do!

My guess would be: by parts.

Here you'd set u = x^2, dv = sinh mx

You'll have an expression in x and sinh mx (or cosh mx) at the end of that, then you'll use parts again, this time with u=x.

I tried integrating by parts with u=sinh (mx) and dV= x^2. Now I have

sinh [mx]/3 - (1/3) integral x^3cosh [mx].

So I tried integrating by parts again, using u= x^3, but it just makes it more complicated.

And sorry my posts are hard to read. Not sure how to use the tags for these.

Originally Posted by veronicak5678

I tried integrating by parts with u=sinh (mx) and dV= x^2. Now I have

sinh [mx]/3 - (1/3) integral x^3cosh [mx].

So I tried integrating by parts again, using u= x^3, but it just makes it more complicated.

And sorry my posts are hard to read. Not sure how to use the tags for these.

of course that makes it more complicated. if you make dv = x^2, you will keep adding a power to the polynomial.

make u = x^2 and dv = sinh(mx). that way, you break down the polynomial each step until it becomes a constant, then the remaining integral is "easy"

Oops. I just misread the first suggestion. Thanks for your help!